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If 25.0 mL of 0.100 M lithium iodide reacts completely with aqueous mercury(II) nitrate,what is the mass of HgI₂ (454.39 g/mol) precipitate? 2 LiI(aq) + Hg(NO₃) 2(aq) → HgI₂(s) + 2 LiNO₃(aq)
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